Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{-4x + 8}{x^2 - 2x - 48} \div \dfrac{3x - 9}{-x^2 + 11x - 24} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-4x + 8}{x^2 - 2x - 48} \times \dfrac{-x^2 + 11x - 24}{3x - 9} $ First factor out any common factors. $k = \dfrac{-4(x - 2)}{x^2 - 2x - 48} \times \dfrac{-(x^2 - 11x + 24)}{3(x - 3)} $ Then factor the quadratic expressions. $k = \dfrac {-4(x - 2)} {(x - 8)(x + 6)} \times \dfrac {-(x - 8)(x - 3)} {3(x - 3)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {-4(x - 2) \times -(x - 8)(x - 3) } { (x - 8)(x + 6) \times 3(x - 3)} $ $k = \dfrac {4(x - 8)(x - 3)(x - 2)} {3(x - 8)(x + 6)(x - 3)} $ Notice that $(x - 8)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {4\cancel{(x - 8)}(x - 3)(x - 2)} {3\cancel{(x - 8)}(x + 6)(x - 3)} $ We are dividing by $x - 8$ , so $x - 8 \neq 0$ Therefore, $x \neq 8$ $k = \dfrac {4\cancel{(x - 8)}\cancel{(x - 3)}(x - 2)} {3\cancel{(x - 8)}(x + 6)\cancel{(x - 3)}} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $k = \dfrac {4(x - 2)} {3(x + 6)} $ $ k = \dfrac{4(x - 2)}{3(x + 6)}; x \neq 8; x \neq 3 $